Unlocking the Secrets of Solving Quadratic Equations Graphically

$f (x) = 2 x^{2} - 2 x - 12$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a > 0 \Rightarrow$ upward	$(\frac{1}{2}, - \frac{2 5}{2})$	$x = \frac{1}{2}$	$(0, - 12)$

f (x) = 2 x^{2} - 2 x - 12

Direction

Vertex

Axis of Symmetry

y -

intercept

a > 0 \Rightarrow

upward

(\frac{1}{2}, - \frac{2 5}{2})

x = \frac{1}{2}

(0, - 12)

Solution	Substitute	Evaluate
$x = - 2$	$f (- 2) = 2 (- 2)^{2} - 2 (- 2) - 12$	$f (- 2) = 0 ✓$
$x = 3$	$f (3) = 2 (3)^{2} - 2 (3) - 12$	$f (3) = 0 ✓$

Solution

Substitute

Evaluate

x = - 2

f (- 2) = 2 (- 2)^{2} - 2 (- 2) - 12

f (- 2) = 0 ✓

x = 3

f (3) = 2 (3)^{2} - 2 (3) - 12

f (3) = 0 ✓

$f (t) = - 16 t^{2} + 112 t - 192$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 4)$	$x = 3.5$	$(0, - 192)$

f (t) = - 16 t^{2} + 112 t - 192

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(3.5, 4)

x = 3.5

(0, - 192)

$g (t) = - 16 t^{2} + 112 t - 196$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(3.5, 0)$	$x = 3.5$	$(0, - 196)$

g (t) = - 16 t^{2} + 112 t - 196

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(3.5, 0)

x = 3.5

(0, - 196)

Area of Square	Area of Triangle
$(0.5 x)^{2}$	$\frac{1}{2} (2) (0.5 x + 2)$
$0.25 x^{2}$	$0.5 x + 2$

Area of Square

Area of Triangle

(0.5 x)^{2}

\frac{1}{2} (2) (0.5 x + 2)

0.25 x^{2}

0.5 x + 2

$y = 0.25 x^{2} - 0.5 x - 2$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a > 0 ⇓ upward$	$(1, - 2.25)$	$x = 1$	$(0, - 2)$

y = 0.25 x^{2} - 0.5 x - 2

Direction

Vertex

Axis of Symmetry

y -

intercept

a > 0 ⇓ upward

(1, - 2.25)

x = 1

(0, - 2)

Area of the Square	Area of the Triangle
$f (x) = (0.5 x)^{2}$	$g (x) = \frac{1}{2} \cdot 2 \cdot (0.5 x + 2)$
$f (x) = 0.25 x^{2}$	$g (x) = 0.5 x + 2$

Area of the Square

Area of the Triangle

f (x) = (0.5 x)^{2}

g (x) = \frac{1}{2} \cdot 2 \cdot (0.5 x + 2)

f (x) = 0.25 x^{2}

g (x) = 0.5 x + 2

$x$	$f (x) = 0.25 x^{2}$	$g (x) = 0.5 x + 2$
$- 3$	$0.25 (- 3)^{2} = 2.25$	$0.5 (- 3) + 2 = 0.5$
$0$	$0.25 (0)^{2} = 0$	$0.5 (0) + 2 = 2$
$3$	$0.25 (3)^{2} = 2.25$	$0.5 (3) + 2 = 3.5$

x

f (x) = 0.25 x^{2}

g (x) = 0.5 x + 2

- 3

0.25 (- 3)^{2} = 2.25

0.5 (- 3) + 2 = 0.5

0

0.25 (0)^{2} = 0

0.5 (0) + 2 = 2

3

0.25 (3)^{2} = 2.25

0.5 (3) + 2 = 3.5

$y = - 0.7 x^{2} + 0.8 x + 8$
Direction	Vertex	Axis of Symmetry	$y -$ intercept
$a < 0 ⇓ downward$	$(\frac{4}{7}, \frac{2 8 8}{3 5})$	$x = \frac{4}{7}$	$(0, 8)$

y = - 0.7 x^{2} + 0.8 x + 8

Direction

Vertex

Axis of Symmetry

y -

intercept

a < 0 ⇓ downward

(\frac{4}{7}, \frac{2 8 8}{3 5})

x = \frac{4}{7}

(0, 8)

Use the graph to solve the quadratic equation. When entering the answer, use the labels to remove or add to the number of roots being entered.

Equation : x^{2} + 8 x + 7 = 0

Equation : - x^{2} - 8 x - 12 = 0

To determine the solutions of the quadratic equation, we need to know the graph of its related function. Quadratic Equation x^2+8x+7=0 ⇓ Related Function y = x^2+8x+7 The solutions of the equation are the x-coordinates of the zeros of function. Since we are given the graph of the related function, we can find them.

The graph crosses the x-axis twice, so the equation x^2+8x+7=0 has two solutions. We can see that those solutions are x=- 7 and x=- 1.

We want to use the given graph to determine the solutions of the given quadratic equation. Recall that the solutions of a quadratic equation are the zeros of its related function. Quadratic Equation - x^2-8x-12=0 ⇓ Related Function y = - x^2-8x-12 We are given the graph of the related function.

This graph crosses the x-axis twice, so its related equation - x^2+2x+3=0 has two solutions. We can see that those solutions are x=- 2 and x=-6 .

Solve the equation by graphing.

x^{2} - 4 x + 4 = 0

- x^{2} = 8 x + 28

We are asked to solve the given quadratic equation by graphing. To do so, these three steps can be followed.

Write the equation in standard form, ax^2+bx+c=0.
Graph the related function y=ax^2+bx+c.
Find the x-intercepts, if any.

The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. Let's identify its related function. Equation:&x^2-4x+4=0 Related Function:&y=x^2-4x+4

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=x^2-4x+4 ⇓ y= 1x^2+( -4)x+ 4 We can see that a= 1, b= -4, and c= 4. Now, we will follow four steps to graph the function.

Find the axis of symmetry.
Calculate the vertex.
Identify the y-intercept and its reflection across the axis of symmetry.
Connect the points with a parabola.

Finding the Axis of Symmetry

The axis of symmetry is a vertical line with equation x=- b2 a. Since we already know the values of a and b, we can substitute them into the formula.

The axis of symmetry of the parabola is the vertical line with equation x=2.

Calculating the Vertex

To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2 a, f( - b/2 a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=2. Therefore, the x-coordinate of the vertex is also 2. To find the y-coordinate, we need to substitute 2 for x in the given equation.

We found the y-coordinate, and now we know that the vertex is (2,0).

Identifying the y-intercept and its Reflection

The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Therefore, the point where our graph intercepts the y-axis is (0, 4). Let's plot this point and its reflection across the axis of symmetry.

Connecting the Points

We can now draw the graph of the function. Since a= 1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.

Finding the x-intercepts

Let's identify the x-intercepts of the graph of the related function.

We can see that the parabola intersects the x-axis once. The point of intersection is ( 2,0). Therefore, the equation x^2-4x+4=0 has one solution, x= 2.

We can check our solution by substituting the value into the given equation. If our solution is correct, the final result will show 0 on both sides of the equation. Let’s check x = 2.

We will follow the same steps to solve the given quadratic equation.

Graph the related function y=ax^2+bx+c.
Find the x-intercepts, if any.

The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Let's first write our equation in the standard form.

Now that we have our equation written in the standard form, let's identify the function related to the equation. Equation:&x^2+8x+28=0 Related Function:&y=x^2+8x+28

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=x^2+8x+28 ⇓ y= 1x^2+ 8x+ 28 We can see that a= 1, b= 8, and c= 28. Now we can identify the characteristics of the function.

y= x^2+8x+28
Direction	Vertex	Axis of Symmetry	y-intercept
a > 0 ⇓ upward	(- 4,12)	x=- 4	(0,28)

The reflection of the y-intercept across the axis of symmetry (- 8,28) is also on the parabola. With this information, we can draw the graph.

Finding the x-intercepts

Let's identify the x-intercepts of the graph of the related function.

We can see that the parabola does not intersect the x-axis. Therefore, the equation x^2+8x+28=0 has no solution. This also means that the original equation, - x^2 = 8x+28 has no solution.

Magdalena's kindergarten-aged little brother is learning to play golf.

The given equation models the height

y

in meters of his golf shot.

y = - x^{2} + 6 x

In this equation,

x

is the horizontal distance in meters. How far away does the golf ball land?

We are given an equation that models the height y of a golf shot depending on the horizontal distance x. y=- x^2+6x Because the ball lands when its height is 0 meters, we can substitute 0 for y in our equation. 0=- x^2+6x We will solve this quadratic equation by graphing. There are three steps for solving a quadratic equation by graphing.

Write the equation in standard form, ax^2+bx+c=0.
Graph the related function y=ax^2+bx+c.
Find the x-intercepts of the graph, if any.

The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. We only have to rearrange it! 0=- x^2+6x ⇔ - x^2+6x=0 Let's write the function related to our equation. Equation:&- x^2+6x=0 Related Function:& y=- x^2+6x To graph the function y=- x^2+6x, we begin by determining the values of a, b, and c. y=- x^2+6x ⇓ y= - 1x^2+ 6x+ 0 Therefore, a= - 1, b= 5, and c= 0. Then we identify the axis of symmetry.

The axis of symmetry for this function is x=3. Next, we determine its vertex. The axis of symmetry always intersects the parabola at its vertex. Therefore, x=3 is the x-coordinate of the vertex. We can use this to determine the corresponding y-coordinate.

The vertex of the function is (3,9). Recall that c= 0, so the y-intercept is (0, 0). Finally we plot the axis of symmetry x=3, the vertex (3,9), and the y-intercept (0,0). We also reflect the y-intercept across the axis of symmetry.

Finally, we connect the three points with a smooth curve.

The x-intercepts of the graph are x=0 and x=6. However, x=0 represents the distance of the ball at the beginning of the shot. Therefore, x=6 is the only solution that makes sense in this context. This means that the golf ball lands 6 meters away from the golfer.

The equation models the height

h

of a soccer ball in centimeters after Maya kicks it, where

t

is the time measured in seconds.

h = - 4 t^{2} + 40 t + 44

After how many seconds does the ball hit the ground?

After how many seconds is the ball

108

centimeters above the ground?

We are given an equation that models the height h of a soccer ball depending on the time t. h=- 4t^2+40t+44 We want to determine after how many seconds the ball hits the ground. The ball lands when its height is 0 feet. Therefore, h= 0. 0=- 4t^2+40t+44 We will solve this quadratic equation by graphing. There are three steps to follow when solving a quadratic equation using the graphing method.

Write the equation in standard form, ax^2+bx+c=0.
Graph the related function y=ax^2+bx+c.
Find the x-intercepts, if any.

The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. We only have to rearrange it! 0=- 4t^2+40t+44 ⇓ - 4t^2+40t+44=0 Let's write the related function of our equation. Equation:&- 4t^2+40t+44=0 Related Function:&h(t)=- 4t^2+40t+44

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. h(t)=- 4t^2+40t+44 ⇓ h(t)= - 4t^2+ 40t+ 44 We can see that a= - 4, b= 40, and c= 44. Now we can identify the characteristics of the function.

h(t)= - 4t^2+40t+44
Direction	Vertex (- b/2a, h(- b/2a ))	Axis of Symmetry t = - b/2a	h -intercept (0,c)
a < 0 ⇓ downward	(5,144)	t = 5	(0,44)

The reflection of the h-intercept across the axis of symmetry, (10,44), is also on the parabola. With this information, we can draw the graph.

Finding the t-intercepts

We can see that the graph has two t-intercepts, one negative and one positive.

Recall that t represents the time after the ball is kicked. Therefore, it cannot be negative. Only the positive t-intercept has meaning in this situation. The positive t-intercept is t=11. This means that the ball hits the ground after 11 seconds.

To find the time when the ball is at a height of 108 centimeters, we need to substitute 108 for h into the given quadratic equation. Then, we will write it in standard form.

Now we can write the function related to the equation. Equation: & - 4t^2 + 40t -64 = 0 Related Function: & h(t)=- 4t^2 +40t - 64 We can now identify the characteristics of the function.

h(t)=- 4t^2 + 40t - 64
Direction	Vertex	Axis of Symmetry	h-intercept
a < 0 ⇓ downward	(5,36)	t= 5	(0,- 64)

In addition to the vertex and the h-intercept, the reflection of the h-intercept across the axis of symmetry, which is at (10,- 64), is also on the parabola. With this information, the graph of the function can be drawn.

The parabola intersects the t-axis twice. The points of intersection are ( 2,0) and ( 8,0).

Therefore, the equation - 4t^2 + 40t-64=0 has two solutions, t_1= 2 and t_2= 8. - 16t^2 +112t-192=0 ↙ ↘ t_1 = 2 t_2 = 8 This means that the ball reaches the height of 108 centimeters twice. After 2 seconds and again after 8 seconds.

Consider the graph shown.

How many solutions does the quadratic equation

x^{2} = - 2 x + 8

have?

Which statement(s) is correct about the graph of

y = x^{2} + 2 x - 8

I . II . III . IV . The vertex is the point (- 1, - 9) . The x - intercepts of the parabola are - 4 and 2 . The y - intercept of the parabola is 4 . The axis of symmetry of the parabola is the line x = 2 .

Usually, to solve a quadratic equation by graphing, we rewrite the equation in standard form, graph its related function, and determine its x-intercepts. However, we can also graph each side of the equation. In this case, the solutions of the equation are the x-coordinates of the points of intersection of the graphs. Let's consider the given equation. x^2= - 2x+8 ⇒ lLHS:y= x^2 RHS:y= - 2x+8 The function related to the left-hand side of the equation is y=x^2, and the function related to the right-hand side of the equation is y=- 2x+8. Note that these functions are plotted on the given graph.

Recall that the solutions of the equation are the x-coordinates of the points of intersection of the graphs related to the sides of the equation. Let's identify them!

The graphs intersect twice. Therefore, the equation x^2=- 2x+8 has two solutions. These solutions are x=- 4 and x=2.

We need to describe what we know about the graph of y=x^2+2x-8. Let's start by identifying the values of a, b, and c. These values can tell us a lot about the graph of a quadratic function. y=x^2+2x-8 ⇓ y= 1x^2+ 2x+( - 8) We have a= 1, b= 2, and c= - 8. First of all, since a>0, we know that the parabola opens upwards. Next we will use the values of a and b to identify the axis of symmetry.

The axis of symmetry for this function is x=- 1. Next we can determine the vertex of the parabola. The axis of symmetry always intersects the parabola at its vertex. Therefore, x=- 1 is the x-coordinate of the vertex. We will use this to determine the corresponding y-coordinate.

The vertex of the parabola is (- 1,- 9). Finally recall that c is the y-intercept of the parabola. Therefore, - 8 is the y-intercept of our parabola. This is all that we can tell about the graph based on the values of a, b, and c. Now we will compare the function from Part B with the equation from Part A. Equation:& x^2=- 2x+8 Function:& y=x^2+2x-8 Let's rewrite the equation in standard form.

We can see that the function related to the equation from Part A is the function from Part B! Equation: & x^2+2x-8 = 0 Related Function: & y = x^2+2x-8

The solutions of the equation are the x-intercepts of the parabola. From Part A we know that the equation has two solutions. Thus, the related function must have two x-intercepts. These x-intercepts are x=- 4 and x=1. Let's sum up the information about the graph of y=x^2+2x-8.

The parabola faces upwards.
The axis of symmetry of the parabola is the vertical line x=- 1.
The vertex is the point (- 1,- 9).
The y-intercept of the parabola is y=- 8.
The x-intercepts of the parabola are x=- 4 and x=2.

Therefore, the statements in I and II are correct. To confirm our observations, we have included a graph of the function y=x^2+2x-8.

Solving Quadratic Equations by Graphing

Catch-Up and Review

Finding Solutions of Quadratic Equations

Quadratic Equation

Solving a Quadratic Equation Graphically

Number of Solutions of a Quadratic Equation

Using a Graph to Find the Number of Solutions of a Quadratic Equation

Finding the Time When a Rocket Reaches a Certain Height

Hint

Solution

Modeling Areas of Geometric Figures With Quadratic Equations

Hint

Solution

Modeling Heights of Arrows With Quadratic Equations

Hint

Solution

Finding Solutions of Quadratic Equations

Graphing Each Side of an Equation

Graphing the Related Function

Finding the Axis of Symmetry

Calculating the Vertex

Identifying the y-intercept and its Reflection

Connecting the Points

Finding the x-intercepts

Graphing the Related Function

Finding the x-intercepts

Graphing the Related Function

Finding the t-intercepts

Solving Quadratic Equations by Graphing

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